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ivann1987 [24]
2 years ago
5

"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

Physics
1 answer:
zavuch27 [327]2 years ago
6 0

Answer:

The value of R is 1.72\times10^{11}\ m.

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

We need to calculate this for value of R

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

R=1.89\times10^{11}\ m

So, The nearest option of the value of R is 1.72\times10^{11}\ m

Hence, The value of R is 1.72\times10^{11}\ m.

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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

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m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

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2 years ago
A gas has an initial volume of 24.6 L at a pressure of 1.90 atm and a temperature of 335 K. The pressure of the gas increases to
Tatiana [17]

Answer:

the final temperature of the gas is 785.18 K

Explanation:

The computation of the final temperature of the gas is shown below:

Here we apply the gas law

= PV ÷ T

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T2 = 785.18 K

hence, the final temperature of the gas is 785.18 K

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